Introduction to Binomial, Exponential and Logarithmic Series

If $-\frac{1}{3}<x<1.$ Then
$1+n\left(\frac{2x}{1+x}\right)+\frac{n(n+1)}{2!}{\left(\frac{2x}{1+x}\right)}^2+\dots \dots \dots \dots$
is equal to

Expand ${\left(1+x\right)}^{\frac{5}{6}}$ up to four terms for $\left|x\right|<1$.

Expand ${\left(1+3x\right)}^{-2}$ up to four terms for $\left|x\right|<\frac{1}{3}$.

Expand ${\left(1+x\right)}^{\frac{3}{4}}$ up to four terms for $\left|x\right|<1$.

Expand ${\left(1+x\right)}^{\frac{1}{2}}$ up to four terms for $\left|x\right|<1$.

Expand ${\left(1+x\right)}^{\frac{2}{3}}$ up to four terms for $\left|x\right|<1$.

Expand the expression $(2x-3{)}^6$

Find ${\left(101\right)}^4$

Find an approximation of ${\left(0.99\right)}^5$ using the first three terms of its expansions. [Write the answer up to three decimal places.]

$\frac{1}{9!}+\frac{1}{3!7!}+\frac{1}{5!5!}+\frac{1}{7!3!}+\frac{1}{9!}$ is equal to

The sum to infinite terms of the series :

$\frac{7}{5}\left(1+\frac{1}{{10}^2}+\frac{1.3}{1.2}·\frac{1}{{10}^4}+\frac{1.3.5}{1.2.3}·\frac{1}{{10}^6}+\dots \dots .\infty \right)=$

If $x=\frac{1}{5}+\frac{1.3}{5.10}+\frac{1.3.5}{5.10.15}+\dots ...\infty$ then $3x^2+6x=$

If $t=\frac{4}{5}+\frac{4\times 6}{5\times 10}+\frac{{\displaystyle 4\times 6\times 8}}{{\displaystyle 5\times 10\times 15}}+.................\infty$, then $9t=$

if n is positive integer and $x$ is any non zero real number, then $c_0+c_1\frac{x}{2}+c_2\frac{x^2}{3}+c_3\frac{x^3}{4}+.............+c_n\frac{x^n}{n+1}=$

If $x=\frac{5}{2!3}+\frac{5·7}{3!3^2}+\frac{5·7·9}{4!3^3}+\dots ,$ then $x^2+4x=$

What is the coefficient of $\frac{y^3}{x^8}$ in $(x+y{)}^{-5},$ when $\left|\frac{y}{x}\right|<1?$

If the middle terms of the expansion ${\left(2x-\frac{x^2}{3}\right)}^{11}$ are $T_p$ and $T_q$. Find the value of $p+q$.

Find the middle term (Terms) in the expansion of

${\left(\frac{\mathrm{x}}{\mathrm{a}}-\frac{\mathrm{a}}{\mathrm{x}}\right)}^{10}$

If the Middle term in the expansion of ${\left({\mathrm{a}}^3+\mathrm{ba}\right)}^{28}$ is ${}^m\mathrm{C}_n{\mathrm{a}}^k{\mathrm{b}}^q$, then $m+n+k+q=$ _____

$\frac{1}{{81}^n}-\frac{10}{{81}^n}·{ }^{2n}{C}_1+\frac{{10}^2}{{81}^n}·{ }^{2n}{C}_2-\frac{{10}^3}{{81}^n}{·}^{2n}{C}_3+\dots \dots +\frac{{10}^{2n}}{{81}^n}=$